91影视

Impulse is the change in the momentum of an object.

\(\begin{array}{c}J = {\left( {mv} \right)_{{\rm{final}}}} - {\left( {mv} \right)_{{\rm{initial}}}}\\ = m\Delta v\end{array}\)

It is also defined as the product of the force applied on an object and the time interval during which the force is applied.

\(\begin{array}{c}J = F{t_{{\rm{final}}}} - F{t_{{\rm{initial}}}}\\ = F\Delta t\end{array}\)

The mass of the ball is\(m = 14\;{\rm{g}}\; = 0.014\;{\rm{kg}}\).

The initial height covered by the ball is\({h_{{\rm{down}}}} = 1.5\;{\rm{m}}\).

The final height covered by the ball after bouncing is \({h_{{\rm{up}}}} = 0.85\;{\rm{m}}\).

Assume the positive direction to be upwards such that the initial velocity of the ball is negative and the final velocity after bouncing back is positive.

Suppose the floor is the zero level for the gravitational potential energy.

Firstly, apply the law of conservation of energy during the 鈥榝alling鈥�. The kinetic energy of the ball at the bottom is equal to its potential energy at the top.

\(\begin{array}{c}K{E_{{\rm{bottom}}}} = P{E_{{\rm{top}}}}\\\frac{1}{2}mv_{{\rm{down}}}^2 = mg{h_{{\rm{down}}}}\\{v_{{\rm{down}}}} = - \sqrt {2g{h_{{\rm{down}}}}} \end{array}\) 鈥� (i)

Now, apply the law of conservation of energy during the 鈥榬ising鈥�.

\(\begin{array}{c}K{E_{{\rm{bottom}}}} = P{E_{{\rm{top}}}}\\\frac{1}{2}mv_{{\rm{up}}}^2 = mg{h_{{\rm{up}}}}\\{v_{{\rm{up}}}} = \sqrt {2g{h_{{\rm{up}}}}} \end{array}\) 鈥� (ii)

The change in the momentum of the ball is equal to the total impulse on the ball. Using equations (i) and (ii) you get,

\(\begin{array}{c}J = m\Delta v\\ = m\left( {{v_{{\rm{up}}}} - {v_{{\rm{down}}}}} \right)\\ = m\left( {\sqrt {2g{h_{{\rm{up}}}}} - \left( { - \sqrt {2g{h_{{\rm{down}}}}} } \right)} \right)\\ = m\sqrt {2g} \left( {\sqrt {{h_{{\rm{up}}}}} + \sqrt {{h_{{\rm{down}}}}} } \right)\end{array}\) 鈥� (iii)

Substitute the known numerical values in equation (iii) for the impulse.

\(\begin{array}{c}J = \left( {0.014\;{\rm{kg}}} \right)\sqrt {2\left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)} \left( {\sqrt {0.85\;{\rm{m}}} + \sqrt {1.5\;{\rm{m}}} } \right)\\ = 0.133\;{\rm{kg}}\;{\rm{m/s}}\end{array}\)

Thus, the impulse on the ball after its hits the floor is \(0.133\;{\rm{kg}}\;{\rm{m/s}}\).